ViewHelper is not found - namespaces

I fiddled around with it for more than 8 hours without getting it resolved.
All I want is to call a ViewHelper in a Fluid template. I did that before and I never had a similar problem.
My ViewHelper file is located in
EXTDIR/VendorName/Classes/ViewHelpers/SomeViewHelper.php.
In my Fluid Template I use the namespace
{namespace k=VendorName\Extname\ViewHelpers}.
Somewhere in the template I call the ViewHelper with
{k:some()}.
The ViewHelper script "SomeViewHelper.php" contains the following code:
class SomeViewHelper extends Tx_Fluid_Core_ViewHelper_AbstractViewHelper {
public function initializeArguments() { }
public function render() {
return 7;
}
}
As you can see, the whole thing is quite simple and the expected output on the page should be "7". But calling a page in the frontend produces this error message:
Oops, an error occurred!
Could not analyse class:VendorName\Extname\ViewHelpers\SomeViewHelper maybe not loaded
or no autoloader?"
Any hints on what might be wrong here?
Cheers
Michael

If everything is spelled correct, and even after deleting System/Configuration Cache the ViewHelper doesn't come up (Could not analyse.. / maybe not loaded or no autoloader), try to reinstall the extension in extension manager!

Your path to the viewHelper source file is wrong.
The correct path should be (without vendorname):
EXTDIR/Classes/ViewHelpers/SomeViewHelper.php
You also need to make sure, you use the correct namespace for your viewHelper (if you're on TYPO3 6+, don't use the old Tx_ classes but namespaces).
<?php
namespace VendorName\Extname\ViewHelpers;
class SomeViewHelper extends \TYPO3\CMS\Fluid\Core\ViewHelper\AbstractViewHelper {
public function initializeArguments() { }
public function render() {
return 7;
}
}

The error message seems to indicate that the class is not found.
Without more info, I would probably suspect a typo somewhere (pun not intended).
The following assumes you are using TYPO3 7 and not composer mode:
First, please check if your viewhelper class is autoloaded. This will help to narrow down the problem. On the command line in the htdocs directory: grep SomeViewHelper typo3temp/autoload/autoload_classmap.php This should give you a hit, if the ViewHelper class is included in the autoload file.
If the classes are not autoloaded, you might manually want to do the autoloading: On the command line in the htdocs directory: php typo3/cli_dispatch.phpsh extbase extension:dumpclassloadinginformation
For more information see: https://docs.typo3.org/typo3cms/CoreApiReference/ApiOverview/Autoloading/Index.html. For more in depth info see this: http://insight.helhum.io/post/130876393595/how-to-configure-class-loading-for-extensions-in You can find this functionality of autoloading in the install tool in TYPO3 8, so in that case you would not need to run the command on the command line.
If this does not work either, check the following:
Are you using namespace correctly in the class: <?php namespace VendorName\Extname\ViewHelpers\SomeViewHelper;
Is the path (including Camelcase class name) correct: /Classes/ViewHelpers/SomeViewHelper.php
After that, clear the system cache and reinitiate the autoloading as described above.
In order for the autoloading to be initiated automatically, you might have to update your ext_emconf.php (if you are not using composer mode) or composer.json (if you are using composer mode).

Just some questions:
Did you add the TypoScript Template of the Extension to the Static Includes?
If you're using a unix-alike system, are you sure that your webserver has the permission to read that files?

If TYPO3 is installed in composer mode and your extension is not installed via composer (e.g. a ProviderExtension of FluidTYPO3), you must provide autoload information for your extension in the main composer.json file, as described within TYPO3-composer-documentation:
In Composer Mode all class loading information must be provided by each of the installed extensions or the root package. If TYPO3 extensions are not installed by composer, e.g. because they are directly committed to the root package or a new package is kickstarted, class loading information needs to be provided, otherwise no classes can be loaded for these extensions/ packages.
E.g. if you have a site extension directly committed to your root package, you must include the class loading information in the root package like that:
Drove me crazy to get my ViewHelpers autoloaded. Providing the autoload-information within a composer.json or the ext_emconf.php of the extension and reinstalling it didn't do the trick.

Related

yii2: model relations error can't find another class in ubuntu server 16.04

I have upload my YII2 project to ubuntu 16.04.
My source is no problem when run on localhost on my computer, but when I run it on the server ubuntu 16.04 with network, it has a problem.
The model source can't find another relation model
public function getLokasiAwal()
{
return $this->hasOne(KotaBandara::className(), ['id_kota' => 'lokasi_awal']);
}
and i have error
Class 'backend\models\TypeNonstaf' not found
I have found the solution, I added the following code:
use backend\models\Kotabandara;
On top in model file but, in my source in localhost,
I do not need to add that code
Can someone explain that issue??
As #rob006 pointed out, it appears that you had been working/running your app on a Windows local file system, which is case-preserving, but not case-sensitive.
When you first call upon a namespaced class directly or via the use operator, it passes this full class name as $className into yii\BaseYii\autoload::($className) (Yii2's global class autoloading handler), which in turn attempts to include the corresponding class file, if found.
So, on your Windows local machine, when you use backend\models\KotaBandara, it will find and include any file associated with the corresponding path alias in a case-insensitive manner, thus it will find any of:
#backend/models/KotaBandara.php
#backend/models/Kotabandara.php
#backend/models/kotabandara.php
#backend/models/KoTaBaNdArA.php
There can be only 1 target file with this sequence of paths/characters anyway.
However, when you migrate this code to a Ubuntu system, which is both case-preserving and case-sensitive, there is a distinct difference between KotaBandara.php and kotabandara.php and in fact both files can exist side by side, unlike on Windows.
So, you have to be precise here - on Ubuntu, use backend\models\KotaBandara will trigger the autoloader to find only the file whose path AND case matches, i.e. KotaBandara.php. If you named the file kotabandara.php, it will be found on Windows, but not on Ubuntu!

Undefined class if class is in different namespace

I have a yii2 project set up on phpstorm 8.0.3. My namespace structure is as follows:
backend\
controllers
models
...
frontend\
controllers
models
..
common\
controllers
models
I used composer on the project and have various dependancies installed in the vendor folder.
Whenever I try to use a class that is not in the same base namespace (frontend, backend or common), I get an undefined class error. The odd thing is the namespace of the class is in autocomplete:
Another oddity is that the undefined class error disappears for some classes if the file with the defined class is opened.
It looks like the PHPStorm cache was somehow corrupted. I ended up invalidating the cache by going to File -> Invalidate Caches/Restart. This deleted my history but fixed the undefined class issue.
You must declare in use section nedded class or namespace. For example, if you need to use BaseController from custom\controllers namespace:
<?php
namespace frontend\components;
// Add this line
use common\controllers\BaseController;
class Controller extends BaseController {
}
I had to reset my PHPstorm to fix the issue of "Class not found".
File > Manage IDE Settings > Restore Default Settings...
Be carreful, it rests everything, but it also offers you to back up your current settings. After that, all has been hard reseted and it now works again.
Can be helpful sometimes.
Settings -> Directories
Remove folders from "Excluded"
I think you have some problems with composer autoload.
Try :
composer update
This was happening to me because of my PHP language settings. I found this out by taking out the namespace in one of my classes. The error changed to "Scalar types are only available in PHP 7".
Go to File -> Settings -> Language & Frameworks -> PHP. Change the PHP language level to language version 7 or greater.
Edit:
After doing that, Insert -> Getters and Setters was intermittently generating functions and comments as mixed instead of string. I don't know why this happened but I recreated all my fields manually and that fixed the issue.

Using CakePHP 3.0 plugin

I'm currently building a new CakePHP app with version 3.0.0-RC1, and trying to install and use the jasig/phpCAS plugin. Using this guide, I've run the following command from the command prompt: composer require jasig/phpcas
This correctly copies the jasig/phpcas files into the vendor directory of my app, but one of the other files that the guide says should be updated (vendor/cakephp-plugins.php) doesn't even exist.
I've had a tough time accessing the plugin. I want to be able to call its static methods, and I keep getting errors of the form: Error: Class 'App\Controller\phpCAS' not found. (The exact directory in the error changes depending on where I'm calling the method from.)
I don't know if this is due to not having the cakephp-plugins.php file, or if I'm not calling the plugin correctly. It's my understanding that if the plugin is loaded I should just be able to call static methods on it like this: phpCAS::methodName()
First of all jasig/phpcas is not a CakePHP plugin. And the vendor/cakephp-plugins.php file is created by the CakePHP plugin installer, so if you don't see such a file, you seem to have either not installed any plugins yet, or you are not using a recent version of the installer, as the creation of this file has been introduced just recently.
Regarding the error about the class not being found, you are missing the leading namespace separator (\phpCAS::methodName()) to access the class in the global namespace, respectively you are missing a proper import (use phpCAS;) that would make the class available in the current namespace.
In case you are not familiar with namespaces, you may want to start with: http://php.net/namespaces

How do I package a log4j configuration file in a NetBeans Platorm application?

Packaging a log4j configuration file in a NetBeans Platform application apparently requires some thinking through. This is what I tried...
I put log4j.xml in src/main/resources/my/package/log4j.xml of some_netbeans_module. The package is a public module package (i.e. classes from this package are used from other packages). I rebuilt the module and confirmed that the file does, in fact, get packaged into the module.
In my classes I get an instance of the logger the way I always do:
static final Logger log = Logger.getLogger(ThisClass.class);
Every NetBeans Platform application has a my_app.conf file which makes it possible to set certain properties. This is where I set log4j.conf:
log4j.configuration="/my/package/log4j.xml"
Now, when I run the application, I see the following output:
[INFO] /home/me/my_app/application/target/my_app/bin/../etc/my_app.conf: 5:
log4j.configuration=/my/package/log4j.xml: not found
What is wrong with the above configuration?
In the my_app.conf file if you append the log4j.configuration property to the default_options property, like so:
default_options="...<other options> -J-Dlog4j.configuration=my/package/log4j.xml"
then this option will get passed to the JVM. Notice that the log4j property has -J-D appended to it. The -J is used by NetBeans to delineate JVM properties and the -D is used by the JVM to delineate a system property.
Also you can/should drop the quotes and the initial / as the quotes are not necessary and NetBeans will complain if you have the initial /
The other way to do this, and the way that I prefer since it doesn't require editing the .conf file, is to put the log4j.xml file into the default package. If you have other requirements that prevents you from doing this then remember that you must put the log4j.configuration property in the app's platform.properties file while your in dev mode and running the app inside of the IDE. Like so:
run.args.extra=-J-Dlog4j.configuration=my/package/log4j.xml
Edit: For questions regarding NetBeans Platform you might have better luck posting to the NetBeans Platform Users forum.

Get the application's path

I've recently searched how I could get the application's directory in Java. I've finally found the answer but I've needed surprisingly long because searching for such a generic term isn't easy. I think it would be a good idea to compile a list of how to achieve this in multiple languages.
Feel free to up/downvote if you (don't) like the idea and please contribute if you like it.
Clarification:
There's a fine distinction between the directory that contains the executable file and the current working directory (given by pwd under Unix). I was originally interested in the former but feel free to post methods for determining the latter as well (clarifying which one you mean).
In Java the calls
System.getProperty("user.dir")
and
new java.io.File(".").getAbsolutePath();
return the current working directory.
The call to
getClass().getProtectionDomain().getCodeSource().getLocation().getPath();
returns the path to the JAR file containing the current class, or the CLASSPATH element (path) that yielded the current class if you're running directly from the filesystem.
Example:
Your application is located at
C:\MyJar.jar
Open the shell (cmd.exe) and cd to C:\test\subdirectory.
Start the application using the command java -jar C:\MyJar.jar.
The first two calls return 'C:\test\subdirectory'; the third call returns 'C:\MyJar.jar'.
When running from a filesystem rather than a JAR file, the result will be the path to the root of the generated class files, for instance
c:\eclipse\workspaces\YourProject\bin\
The path does not include the package directories for the generated class files.
A complete example to get the application directory without .jar file name, or the corresponding path to the class files if running directly from the filesystem (e.g. when debugging):
String applicationDir = getClass().getProtectionDomain().getCodeSource().getLocation().getPath();
if (applicationDir.endsWith(".jar"))
{
applicationDir = new File(applicationDir).getParent();
}
// else we already have the correct answer
In .NET (C#, VB, …), you can query the current Assembly instance for its Location. However, this has the executable's file name appended. The following code sanitizes the path (using System.IO and using System.Reflection):
Directory.GetParent(Assembly.GetExecutingAssembly().Location)
Alternatively, you can use the information provided by AppDomain to search for referenced assemblies:
System.AppDomain.CurrentDomain.BaseDirectory
VB allows another shortcut via the My namespace:
My.Application.Info.DirectoryPath
In Windows, use the WinAPI function GetModuleFileName(). Pass in NULL for the module handle to get the path for the current module.
Python
path = os.path.dirname(__file__)
That gets the path of the current module.
Objective-C Cocoa (Mac OS X, I don't know for iPhone specificities):
NSString * applicationPath = [[NSBundle mainBundle] bundlePath];
In Java, there are two ways to find the application's path. One is to employ System.getProperty:
System.getProperty("user.dir");
Another possibility is the use of java.io.File:
new java.io.File("").getAbsolutePath();
Yet another possibilty uses reflection:
getClass().getProtectionDomain().getCodeSource().getLocation().getPath();
In VB6, you can get the application path using the App.Path property.
Note that this will not have a trailing \ EXCEPT when the application is in the root of the drive.
In the IDE:
?App.Path
C:\Program Files\Microsoft Visual Studio\VB98
In .Net you can use
System.IO.Directory.GetCurrentDirectory
to get the current working directory of the application, and in VB.NET specifically you can use
My.Application.Info.DirectoryPath
to get the directory of the exe.
Delphi
In Windows applications:
Unit Forms;
path := ExtractFilePath(Application.ExeName);
In console applications:
Independent of language, the first command line parameter is the fully qualified executable name:
Unit System;
path := ExtractFilePath(ParamStr(0));
Libc
In *nix type environment (also Cygwin in Windows):
#include <unistd.h>
char *getcwd(char *buf, size_t size);
char *getwd(char *buf); //deprecated
char *get_current_dir_name(void);
See man page
Unix
In unix one can find the path to the executable that was started using the environment variables. It is not necessarily an absolute path, so you would need to combine the current working directory (in the shell: pwd) and/or PATH variable with the value of the 0'th element of the environment.
The value is limited in unix though, as the executable can for example be called through a symbolic link, and only the initial link is used for the environment variable. In general applications on unix are not very robust if they use this for any interesting thing (such as loading resources). On unix, it is common to use hard-coded locations for things, for example a configuration file in /etc where the resource locations are specified.
In bash, the 'pwd' command returns the current working directory.
In PHP :
<?php
echo __DIR__; //same as dirname(__FILE__). will return the directory of the running script
echo $_SERVER["DOCUMENT_ROOT"]; // will return the document root directory under which the current script is executing, as defined in the server's configuration file.
echo getcwd(); //will return the current working directory (it may differ from the current script location).
?>
in Android its
getApplicationInfo().dataDir;
to get SD card, I use
Environment.getExternalStorageDirectory();
Environment.getExternalStoragePublicDirectory(String type);
where the latter is used to store a specific type of file (Audio / Movies etc). You have constants for these strings in Environment class.
Basically, for anything to with app use ApplicationInfo class and for anything to do with data in SD card / External Directory using Environment class.
Docs :
ApplicationInfo ,
Environment
In Tcl
Path of current script:
set path [info script]
Tcl shell path:
set path [info nameofexecutable]
If you need the directory of any of these, do:
set dir [file dirname $path]
Get current (working) directory:
set dir [pwd]
Java:
On all systems (Windows, Linux, Mac OS X) works for me only this:
public static File getApplicationDir()
{
URL url = ClassLoader.getSystemClassLoader().getResource(".");
File applicationDir = null;
try {
applicationDir = new File(url.toURI());
} catch(URISyntaxException e) {
applicationDir = new File(url.getPath());
}
return applicationDir;
}
in Ruby, the following snippet returns the path of the current source file:
path = File.dirname(__FILE__)
In CFML there are two functions for accessing the path of a script:
getBaseTemplatePath()
getCurrentTemplatePath()
Calling getBaseTemplatePath returns the path of the 'base' script - i.e. the one that was requested by the web server.
Calling getCurrentTemplatePath returns the path of the current script - i.e. the one that is currently executing.
Both paths are absolute and contain the full directory+filename of the script.
To determine just the directory, use the function getDirectoryFromPath( ... ) on the results.
So, to determine the directory location of an application, you could do:
<cfset Application.Paths.Root = getDirectoryFromPath( getCurrentTemplatePath() ) />
Inside of the onApplicationStart event for your Application.cfc
To determine the path where the app server running your CFML engine is at, you can access shell commands with cfexecute, so (bearing in mind above discussions on pwd/etc) you can do:
Unix:
<cfexecute name="pwd"/>
for Windows, create a pwd.bat containing text #cd, then:
<cfexecute name="C:\docume~1\myuser\pwd.bat"/>
(Use the variable attribute of cfexecute to store the value instead of outputting to screen.)
In cmd (the Microsoft command line shell)
You can get the name of the script with %* (may be relative to pwd)
This gets directory of script:
set oldpwd=%cd%
cd %0\..
set app_dir=%pwd%
cd %oldpwd%
If you find any bugs, which you will. Then please fix or comment.
I released https://github.com/gpakosz/whereami which solves the problem in C and gives you:
the path to the current executable
the path to the current module (differs from path to executable when calling from a shared library).
It uses GetModuleFileNameW on Windows, parses /proc/self/maps on Linux and Android and uses _NSGetExecutablePath or dladdr on Mac and iOS.
Note to answer "20 above regarding Mac OSX only: If a JAR executable is transformed to an "app" via the OSX JAR BUNDLER, then the getClass().getProtectionDomain().getCodeSource().getLocation(); will NOT return the current directory of the app, but will add the internal directory structure of the app to the response. This internal structure of an app is /theCurrentFolderWhereTheAppReside/Contents/Resources/Java/yourfile
Perhaps this is a little bug in Java. Anyway, one must use method one or two to get the correct answer, and both will deliver the correct answer even if the app is started e.g. via a shortcut located in a different folder or on the desktop.
carl
SoundPimp.com